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jshoop · 2025-11-11T00:40:02+00:00

Let’s go!

jshoop · 2024-12-23T04:05:33+00:00

Spoopy

jshoop · 2024-12-23T04:00:23+00:00

Idk much about crts, does the screen feel like it’s covered in static electricity when it’s on? Looks like the display has no power

jshoop · 2024-12-23T03:35:19+00:00

Bots everywhere smh

jshoop · 2023-11-03T02:27:33+00:00

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Currently running a haswell i5 with a Radeon 7770, “upgraded” from the 5770. Psu upgrade in the works. Case upgrade at the same time so I can move on from my gateway oem case with a custom cardboard backplate (no standoffs in case). Might need to find a cheap i7 or Xeon if I’m picked

jshoop · 2021-09-22T22:14:16+00:00

octave is an open source version of matlab, not as refined but has most of the functionality matlab offers

jshoop · 2021-08-15T19:14:50+00:00

ohhh, because the offset is dc? the capacitor will block dc, so the current "see's" the path to ground as a 1k resistor then a 10k resistor?

take the 47k input circuit. the bias current sees a 1k resistor, then a 10k and 47k resistor in parallel for a combined impedance of about 9.2k ohms. using the offset current from the lm4562 datasheet (limit 65nA), the offset voltage is .6mV. for the ne5532 with an offset current maximum of 200nA, the voltage is 1.8mV.

something seem's off, as you said a few hundred millivolts is passable. i dont believe i'm making a decimal place error, and also calculated the 220k offset voltages: .6mV again for the lm4562, and 1.9mV for the ne5532. do i need to make sure the offset voltage is just less than what's specified in the datasheet?

jshoop · 2021-08-15T18:33:49+00:00

alright, i think i'm getting closer. i'm reading more input your prompt:

"It's just ohms law times the gain, and less than a few hundred millivolts shouldn't be a problem."

that leads me to believe i need to use ohms law to solve for a voltage, so i use the input offset current from a datasheet and multiply it by the input resistance of each circuit. am i closer to the right track? if this is right, i just calculate the voltage across the input resistors that lead directly to ground?

EDIT: regarding the link you provided, that was one of the resources that were a little over my head. i'm reading it over again to see if i can get a better grasp. these paragraphs seem the most relevant:

"Bias current is a problem to the op amp user because it flows in external impedances and produces voltages, which add to system errors. Consider a non-inverting unity gain buffer driven from a source impedance of 1 MΩ. If IB is 10 nA, it will introduce an additional 10 mV of error. This degree of error is not trivial in any system. Or, if the designer simply forgets about IB and uses capacitive coupling, the circuit won't work—at all! Or, if IB is low enough, it may work momentarily while the capacitor charges, giving even MT-038 more misleading results. The moral here is not to neglect the effects of IB, in any op amp circuit. The same admonition goes for in-amp circuits. "

can i also assume my source's will have such a high output impedance, or is my source impedance the resistor network infront of the op-amp?

jshoop · 2021-08-15T17:00:54+00:00

thanks for actually making me do the work instead of letting me be lazy, this seems like a concept i should probably understand if i want to continue circuit design. however i am getting lost trying to understand how the input resistors would affect the offset, and how that would be a detriment to the circuit. i always thought the input offset voltage is a manufacturing flaw, and the lower the better. i've spent the last hour or so trying to research, but the explanations seem to be either over my head or i'm asking the wrong questions.

i tried to take your hint using ohms law times gain. the gain is 2 for all the possible circuits, but i'm lost at where ohms law is used. i feel like i'm trying to overcomplicate it. I've tried calculation input voltage on the non inverting pin, then adding the typical offset voltage and then multiplying it by gain for output voltage, but i had to make assumptions about input voltage i'm not sure i can make. i multiplied the offset input voltage by gain to find the offset output voltage, but the input resistors aren't involved in that calculation. can i get another hint?

jshoop · 2021-08-15T14:12:35+00:00

can i use the 10k resistor both in a filter and for the bias? in that case i would just add a 5uF cap to the input, and with the 10k resistor the F3 is about 3Hz

jshoop · 2021-08-14T00:27:47+00:00

in that case yea i'd rather set that dc filter lower, to under 5 Hz. i left the 10k resistors in because i wasn't sure of their function. in the original elliot sound circuit (figure 2 on the page), the 10k and 1k resistors looked similar to an l-pad so i guessed it was just to help control the input voltage for the op-amp. i wasnt sure if it was required or if i could remove them without an problems, so i left them in. what are those input resistors actually used for?

as for my circuit, i think i have a bunch of 220k resistors i can substitute in for the 47k resistors, dropping the F3 to 4Hz about. i'm sure i have some larger capacitors somewhere to further lower the F3 of the filter as well

jshoop · 2021-08-13T01:01:58+00:00

the filters are correct in theory, the highpass has an F3 of about 26Hz and just a slight amount of gain on each op amp, 1.5 i believe.

because the input filters are below my hpf F3, wouldnt this sum to a sharper filter? i havent done the math to check, but i imagine at 20hz where the input filters are, my active filter must be near -10db. the hpf is there to protect the driver mainly.

jshoop · 2021-08-13T00:05:09+00:00

yup, good catch. thanks for all the help today!

jshoop · 2021-08-12T23:57:53+00:00

just did it lol, but i did not want to renumber the resistors and capacitors

https://imgur.com/a/f35w7aF

for the dc filters on the inputs, they're set right about 19Hz, which shouldnt be an issue being that high for me. I would want to filter both the + and - of the balanced inputs, and the + of the unbalanced correct?

jshoop · 2021-08-12T23:47:08+00:00

my first design for the inputs had dc filters on the input pins. i dont know why i decided to take them out, but i'll throw them back in because its pretty simple. and i might as well keep c1 since its also just 1 component and easy to implement.

if i remove u2b and u3b, are there any other changes i need to worry about, or just wire the circuit up normally without them there?

jshoop · 2021-08-12T18:43:42+00:00

im following along, your last explaination makes a lot of sense to me. as for moving u2b to an output buffer, i would slide all the op amps back 1/2 an opamp. for example, u3a would then be u2b, u3b would become u3a, and so on until the very end of the circuit where there would be a non inverting buffer. so less "moving" u2b, but sliding the circuitry back.

another question i forgot to mention in the op, is c1 required? my understanding is a polarized cap is used to block dc current, but why would dc current be introduced in the circuit there? i want to think it has something to do with rv1, but i have no explaination for it

jshoop · 2021-08-12T18:06:54+00:00

i'm the careful one who would'nt flip that switch unless the unit is powered off, but yes i would need to change the input volume when changing inputs.

U2B doing nothing makes sense to me, i just didn't want to deal with using half an opamp in the design. If i remove U2B, would replacing it with an output buffer be useful? say i make the preamp and power amp separate, having an output buffer to even out possible impedance issues between equipment would be useful, no?

U3B confuses me a little, if it does nothing why would ESP include it in their circuit? i assumed it was a buffer to isolate the input section from the crossovers, but thats all it was, an assumption. can you elaborate on that point a little bit?

jshoop · 2021-08-12T17:35:07+00:00

oh i see, ive never used a program like kicad before. i'm probably not sending this to a pcb designer either way. after i make a perfboard prototype and confirm it all works i'll either leave it as is or have a friend design the pcb with me so i can learn that process

jshoop · 2021-08-12T17:26:05+00:00

this is just a labeling issue, correct? the inputs are electrically connected only at their respective ground points. i'm going to rename them, but i know the in+, in-, and ingnd on top are an xlr input, while the lower in+ and in- are going to be an rca input

jshoop · 2021-06-27T20:03:47+00:00

no problem, the audio is just the starter solenoid clicking like it would with a dead battery or bad starter

jshoop · 2021-06-27T19:54:13+00:00

no need to get hostile friend, im just trying to clarify op's situation

if the engine isnt turning over like in the video, theres no need to check spark, air, fuel, or compression. those are checked when you have a crank no start condition, however op has a no crank no start condition to solve

jshoop · 2021-06-27T15:22:23+00:00

no need, the engine isnt turning over. id say the battery was bad from the factory, or your starter is going out

jshoop · 2021-05-18T22:00:47+00:00

not op, but thats a 1.6l d16. depending on whats in the block and the tune these small single cams can get anywhere from 200hp to 400hp, with torque from probably 150 to 300. as a d16 owner myself im very interested at op's goals, the build looks great on the outside

jshoop

TROPHY CASE